What is the integral of ((1-cos(4x))/2)^2 ?

The integral of ((1-cos(4x))/2)^2 is 1/2 (-1/4 sin^3(2x)cos(2x)+3/8 (2x-1/2 sin(4x)))+C

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integral of ((1-cos(4x))/2)^2

\int (\frac{1 - cos ( 4 x )}{2})^{2} d x

\frac{1}{2} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x))) + C

Solution steps

\int (\frac{1 - cos ( 4 x )}{2})^{2} d x

Rewrite using trig identities

= \int sin^{4} (2 x) d x

Apply u-substitution

: \int sin^{4} (u) \frac{1}{2} d u

= \int sin^{4} (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

\int sin^{2} (u) d u = \frac{1}{2} (u - \frac{1}{2} sin (2 u))

= \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 2 x

= \frac{1}{2} (- \frac{cos ( 2 x ) sin ^{3} ( 2 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 x - \frac{1}{2} sin (2 \cdot 2 x)))

Simplify

\frac{1}{2} (- \frac{cos ( 2 x ) sin ^{3} ( 2 x )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 x - \frac{1}{2} sin (2 \cdot 2 x))) : \frac{1}{2} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x)))

= \frac{1}{2} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x)))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4} sin^{3} (2 x) cos (2 x) + \frac{3}{8} (2 x - \frac{1}{2} sin (4 x))) + C

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What is the integral of ((1-cos(4x))/2)^2 ?
The integral of ((1-cos(4x))/2)^2 is 1/2 (-1/4 sin^3(2x)cos(2x)+3/8 (2x-1/2 sin(4x)))+C